Section 8.2
The basic normal-shock equations
A normal shock stands perpendicular to the flow, for example on the stagnation streamline of the bow shock ahead of a blunt body, inside the diverging section of a nozzle (see the Chapter 10 lesson), or at the inlet of a supersonic engine. The problem is simply stated: given the flow ahead of the wave (p1, ρ1, T1, M1, u1, …), find the flow behind it (p2, ρ2, T2, M2, u2, …).
The shock itself is a thin region of highly viscous flow with steep gradients of velocity and temperature, so the flow through it is adiabatic but not isentropic. We avoid its interior entirely by drawing a control volume abcd whose faces ab and cd lie in the uniform flow on either side. The analysis assumes that
- the flow is steady,
; - the flow is adiabatic,
: no heat is added to or taken from the control volume; - there are no viscous stresses on the faces ab and cd, which lie outside the shock;
- there are no body forces,
.
Evaluate the steady integral conservation laws over abcd. Only faces ab and cd carry flow; the sides bc and da are parallel to the velocity, and the pressure forces on them have no x component. With uniform properties of area A on each face, the three laws reduce to algebra:
The energy result uses
Counting equations and unknowns. With region 1 known, continuity, momentum and energy are three equations in four unknowns, ρ2, u2, p2 and h2. Two thermodynamic relations close the system, each adding one unknown, T2:
| Equation | Form | Equations | Unknowns |
|---|---|---|---|
| Continuity | ρ1u1 = ρ2u2 | 1 | ρ2, u2 |
| Momentum | p1 + ρ1u1² = p2 + ρ2u2² | 1 | + p2 |
| Energy | h1 + u1²/2 = h2 + u2²/2 | 1 | + h2 |
| Enthalpy | h2 = cpT2 | 1 | + T2 |
| Equation of state | p2 = ρ2RT2 | 1 | — |
| Total | 5 | 5 unknowns |
Section 8.3
Speed of sound
Ride with a sound wave, so that the wave is stationary and the gas enters it at speed a. Across the wave the properties change by infinitesimal amounts. Gradients are weak, so viscosity and heat conduction are negligible: the flow through a sound wave is adiabatic and reversible, that is, isentropic. Continuity (8.2) and momentum (8.6) across the wave give
Eliminating da leaves a² = dp/dρ, and since the change is isentropic,
For a calorically perfect gas the isentropic relation p = cργ gives (∂p/∂ρ)s = γp/ρ, and with p = ρRT,
The speed of sound of a calorically perfect gas depends on temperature only. At standard sea level, T = 288.16 K and a = 340 m/s.
In terms of the isentropic compressibility
The less compressible the gas, the faster sound travels. An incompressible fluid (τs = 0) has an infinite speed of sound, so the incompressible flows of Chapters 3–6 are, in theory, flows at zero Mach number.
What the Mach number measures. The ratio of the kinetic energy per unit mass to the internal energy per unit mass, with e = cvT and cv = R/(γ − 1), is
M² measures the directed (kinetic) motion of the gas against the random thermal motion of its molecules. At M = 2 the two are comparable (ratio 1.12); at M = 20 the kinetic energy is 112 times the internal energy, which is why hypersonic flows heat so strongly when they are brought to rest.
Standard atmosphere of Anderson’s Appendix D: T falls 6.5 K/km to 11 km geopotential altitude, stays at 216.66 K to 25 km, then rises 3 K/km. Geometric altitude is converted to geopotential altitude; the values match Appendix D to within 0.2%.
Worked examples 8.1–8.6
An airplane flies at m/s. Calculate its Mach number at standard altitudes of (a) sea level, (b) 5 km and (c) 10 km.
The temperature falls with altitude, so the speed of sound falls and the same airspeed is a higher Mach number. Sea level uses T = 288 K in the notes (288.16 K in Appendix D).
At a point in the flow of Example 7.3 the temperature is K and the velocity is m/s. Calculate the Mach number at this point.
This is a local Mach number: it varies from point to point in the flow, unlike the freestream Mach number of Example 8.1.
Calculate the ratio of kinetic energy to internal energy at a point in an airflow where the Mach number is (a) and (b) .
At a point in a flow of air the pressure and density are atm and kg/m³. (a) Calculate the isentropic compressibility. (b) From it, calculate the speed of sound.
For an isentropic process p/ργ = c, and differentiating v = 1/ρ = (p/c)−1/γ gives τs = 1/(γp). The notes convert 1 atm as 1.01 × 105 N/m² (exact: 1.01325 × 105), hence τs = 1.0103 × 10−5 m²/N there. The density appears as 0.978 kg/m³ in the statement and 0.784 kg/m³ in the check; the numbers in the solution correspond to about 0.79 kg/m³, the value used here.
A tube m long is filled with gas at K. A sound wave is generated at one end. How long does it take to reach the other end if the tube is filled with (a) air, (b) helium (γ = 1.67, R = 2078.5 J/(kg·K))?
At the same temperature, sound travels faster in helium because its gas constant is 7.2 times larger (its molecules are lighter) and its γ is higher. Example 8.5 is left as an exercise in the notes.
Section 8.4
Special forms of the energy equation
For steady, adiabatic, inviscid flow, h + u²/2 is constant along a streamline (8.29). For a calorically perfect gas, h = cpT and cp = γR/(γ − 1), so
Two reference states turn (8.32) into a constant. Bring the flow to rest (u = 0, a = a0), or accelerate or decelerate it to sonic speed (u = a = a*):
so that
Total conditions. With u2 = 0 in (8.30), cpT + u²/2 = cpT0 (8.38), and dividing by cpT:
Total pressure and density are defined by an isentropic compression to rest,
For a given gas these ratios depend on the Mach number only; Appendix A tabulates them for γ = 1.4.
Sonic conditions. Setting M = 1 in (8.40), (8.42) and (8.43):
For air (γ = 1.4) these are 0.833, 0.528 and 0.634: numbers worth remembering.
Characteristic Mach number. Define M* = u/a*, the velocity measured against the sonic speed of sound a* = √(γRT*) rather than the local one. Dividing (8.35) by u² and solving:
M* < 1 when M < 1, M* = 1 when M = 1 and M* > 1 when M > 1, so M* behaves like M qualitatively. But as M → ∞, M* tends to a finite limit,
Worked examples 8.8–8.10
At a point in an airflow the local Mach number, static pressure and static temperature are , atm and K. Calculate the local p0, T0, T*, a* and M*.
p0 and T0 are the conditions the gas would reach if brought to rest isentropically; T* and a* those it would reach if brought isentropically to M = 1. M* follows either from u/a* or directly from (8.48).
An airfoil is in a freestream with M∞ = , p∞ = atm and T∞ = K. At point 1 on the airfoil the pressure is p1 = atm. Assuming isentropic flow, calculate the local Mach number (Ex 8.9) and velocity (Ex 8.10) at point 1.
The flow outside the boundary layer is isentropic, so p0 is the same at every point: p0,1 = p0,∞. The local p0/p then fixes the local Mach number. The temperature at point 1 follows from the isentropic relation T1/T∞ = (p1/p∞)(γ−1)/γ.
Section 8.6
Calculation of normal-shock properties
Return to the five equations of Section 8.2. Divide momentum (8.6) by continuity (8.2) and use a² = γp/ρ:
The energy equation in the form (8.35) holds on both sides with the same a*, because h0 is unchanged across the shock:
Derivation of the Prandtl relation (three lines)
Dividing (8.55) by a*² gives
Continuity and the Prandtl relation give the density ratio,
Momentum and continuity give
The equation of state then gives the temperature ratio, and h = cpT the enthalpy ratio:
Limits. With M1 = 1 all ratios equal 1 and M2 = 1: a normal shock of vanishing strength is a Mach wave. As M1 increases the shock strengthens and M2 falls below 1, so the flow behind a normal shock is always subsonic. As M1 → ∞ (γ = 1.4):
Section 8.6 · The second law
Entropy and total pressure across the shock
The relations above embody only continuity, momentum and energy, which do not care whether M1 is subsonic or supersonic. Mathematically they also return a “shock” for M1 < 1, an expansion in which the pressure drops. Only the second law of thermodynamics decides which direction the process can take. The entropy change of a calorically perfect gas, with (8.65) and (8.67), is
For M1 = 1 the change is zero; for M1 > 1 it is positive; for M1 < 1 it is negative, which the second law forbids. Normal shocks occur only in supersonic flow; an expansion shock is impossible.
Nature produces the entropy rise inside the wave. The shock is a region roughly 10−5 cm thick (a few mean free paths at sea-level density), and across it the large gradients of velocity and temperature make friction and heat conduction strong. Both are dissipative, irreversible mechanisms that always increase entropy.
Total conditions on each side. Imagine bringing a fluid element ahead of the shock to rest isentropically (state 1a: p0,1, T0,1, s1), and an element behind the shock likewise (state 2a: p0,2, T0,2, s2). The flow through the shock is adiabatic, so the energy equation gives
Writing the entropy change between the two rest states, the temperature term vanishes:
Since s2 > s1, total pressure always falls across a shock, while total temperature stays constant. The loss of p0 is the measure of the shock’s irreversibility; Appendix B lists p0,2/p0,1.
Worked examples
Normal shocks: Examples 8.11–8.21
The Result column comes from the relations above. Change any given value and it updates. Switch on Compare with the hand solution to add the solution from Appendices A and B (γ = 1.4, original values); a † marks a value that differs from the result, and the note under the table explains why.
A normal shock wave stands in air where the upstream properties are u1 = m/s, T1 = K and p1 = atm. Calculate the velocity, temperature and pressure downstream of the shock.
u1 = 680 m/s gives M1 = 1.999, which the notes round to 2 before entering Appendix B. The velocity behind the shock also follows directly from continuity: u2 = u1/(ρ2/ρ1).
A normal shock stands in a supersonic airstream where the pressure upstream of the shock is atm. Calculate the loss of total pressure across the shock when the upstream Mach number is (a) M1 = and (b) M1 = . Compare the two results.
A ramjet flies at M∞ = at 10 km, where p∞ = N/m² and T∞ = K. A normal shock stands at the inlet (point 1); the inlet then slows the flow isentropically to M2 = at the combustor entrance (point 2). Calculate p2 and T2. Example 8.14 repeats the calculation for M∞ = 10, with the inlet redesigned so that M2 stays 0.2.
Total pressure is lost across the shock and then stays constant through the isentropic inlet; total temperature stays constant throughout. At Mach 10 the air would enter the combustor at about 33 atm and 4650 K, far beyond what any combustor or structure tolerates. This is why hypersonic air-breathing engines keep the flow supersonic through the combustor: the scramjet (supersonic-combustion ramjet), demonstrated near Mach 10 by NASA’s X-43A.
(8.15) The pressure ratio across a normal shock in air is 4.5. (8.16) The temperature ratio across a normal shock in air is 5.8. (8.17) The Mach number behind a normal shock is 0.4752. In each case, find the Mach numbers ahead of and behind the shock and the remaining property ratios.
For a normal shock, any one of p2/p1, T2/T1, ρ2/ρ1, M2 or p0,2/p0,1 fixes M1, and with it every other ratio. Use the “Known quantity” box of Interactive 3 to invert any of them.
(8.18) The velocity and temperature ahead of a normal shock are u1 = m/s and T1 = K. Calculate the velocity behind the shock. (8.19) The velocity and temperature behind a normal shock are u2 = m/s and T2 = K. Calculate the velocity ahead of the shock. (8.20) Repeat 8.19 using the nearest table entry instead of interpolation.
Unlike Examples 8.15–8.17, neither the velocity nor the temperature alone defines a normal shock: both are needed to find the Mach number. In Example 8.19, M2 = 0.4238 falls between the Appendix B entries M1 = 4.45 (M2 = 0.4245) and 4.50 (0.4236); the notes interpolate linearly. Using the nearest entry instead (Ex 8.20) changes u1 by less than 0.1%.
A normal shock has an upstream Mach number of M1 = (enter a value between 2 and 5). Obtain the downstream Mach number by (a) the nearest entry in Appendix B, (b) linear interpolation between entries and (c) the exact relation (8.59). Compare the three.
Between M1 = 2 and 5, Appendix B steps M1 by 0.05, and the table entries here are those values rounded to four decimals. For practical purposes the three approaches agree; the notes adopt the nearest-entry method for the rest of the course.
Section 8.7
Measurement of velocity in a compressible flow
In low-speed flow, the total pressure from a Pitot tube and the static pressure from a wall orifice give the velocity through Bernoulli’s equation. In compressible flow the same two pressures still give the Mach number, but the formula depends on the Mach-number regime.
Subsonic compressible flow
The mouth of the tube (point b) is a stagnation point, and the gas along streamline ab is brought to rest isentropically. The tube therefore reads the freestream total pressure p0,1, and (8.42) inverts to
Unlike incompressible flow, p0,1 and p1 alone do not give u1: the freestream speed of sound, that is T1, is also needed.
Supersonic flow
Ahead of the tube stands a detached bow shock. A fluid element on the stagnation streamline cde first crosses the normal part of the shock, decelerating non-isentropically to a subsonic speed at d, then is compressed isentropically to rest at e. The tube therefore reads p0,2, the total pressure behind a normal shock, which is less than p0,1. Writing p0,2/p1 = (p0,2/p2)(p2/p1) with the isentropic relation for the first factor, (8.59) for M2 and (8.65) for the second factor gives the Rayleigh Pitot-tube formula:
Appendix B tabulates p0,2/p1 against M1. At M1 = 1 both formulas give 1.893 for air: a reading above 1.893 p1 means the flow is supersonic.
Worked examples 8.22–8.24
A Pitot tube is inserted into an airflow where the static pressure is atm. Calculate the flow Mach number when the tube measures (a) atm, (b) atm and (c) atm.
First decide the regime: at M1 = 1 the tube reads 1.893 p1. Below that, the tube senses the freestream total pressure (Appendix A); above it, the total pressure behind the bow shock (p0,2/p1, Appendix B). The last column shows the Mach number you would get by wrongly applying the isentropic formula to a supersonic reading.
A hypersonic missile flies at Mach at an altitude where the pressure is N/m². Its nose is blunt. Calculate the pressure at the stagnation point on the nose.
The stagnation point lies behind the normal part of the bow shock, so its pressure is p0,2, not the freestream total pressure. The notes state the altitude as 6 km; 4.662 × 104 N/m² is the standard pressure at 20,000 ft (6.1 km), the altitude in Anderson’s original statement. The notes convert with 1 atm = 1.01 × 105 N/m², hence 38.2 atm.
A Lockheed SR-71 Blackbird flies at a standard altitude of 25 km, where Appendix D gives T1 = K and p1 = N/m². The Pitot tube on the airplane measures N/m². Calculate the velocity of the airplane.
Invert the Rayleigh Pitot formula for M1, then u1 = M1a1. As in subsonic flow, the temperature is needed to turn a Mach number into a velocity.
Self-check
Check your understanding
Pick an answer to see whether it is right and why. Use the normal-shock explorer to test your reasoning.
Summary